I have strong suspicion that the following is true, though I am stuck in proving it;
Two matrices $A , B $ are positve semidefinite while $M$ is positive definite, meaning that they are all square, symmetric and have nonnegative eigenvalues, in particular $M$ has all positive eigenvalues. Then $AMB + BMA$ is symmetric, and I think that its eigenvalues are nonnegative. That would make $AMB + BMA$ positive semidefinite, which would make me happy!
Any ideas if it is possible to prove, or disproove?
I have considered using the weak majorization of Hermitian matrices, but I am not able to apply it to the matrix triplet above ...
Furthermore, since $AMB + BMA$ is symmetric, it is unitary diaonalizable to a real diagonal matrix, i.e. equals
$$U^T D U = (U^T D^{1/2}) (D^{1/2}U) $$
thus also being decomposable to the form $P^* P$, thus being positive semidefinite - does this hold? NO, since the matrix $D$ could have negative values, thus not possible to split up into real numbers.
I believe, unfortunately, you are wrong.
Consider the following matrices:
\begin{align} A &= \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} & B &= \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} & M &= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \end{align}
They do follow your requierements but: \begin{equation} C = AMB + BMA = \begin{pmatrix} 2 & 1 \\ 1 & 0 \end{pmatrix} \end{equation} is not PSD. Its determinent is negative $\det C = -1$ which means that both eigenvalues have of opposite signs.